diketahui f(x-3) = (x^2+3x)/(x^2+7x) tentukan limit f(x) dmn x mendekati 3

[tex]f(x-3)=\frac{x^2 + 3x}{x^2 + 7x} \\ f(x-3)=\frac{x+3}{x+7} \\ f(x-3+3) = \frac {x+3+3}{x+3+7} \\ f(x) = \frac {x+6}{x+10} \\ \lim_{x \rightarrow 3} \frac{x+6}{x+10} =\frac {3+6}{3+10} = \frac {9}{10}[/tex]